Show that $\Bbb Z[\sqrt{-5}]$ satisfies ascending chain condition for principal ideals

Question

I have got this question from my friend now he is at the beginning of algebra. I can prove this using $\Bbb Z[x]$ Noetherian and so is its quotient but it uses techniques of field extension(quotients) and Noetherian domain. He is just learning PID, UFD, ACCP and all so is there any direct technique starting that $\langle a_1\rangle\subseteq \langle a_2\rangle \subseteq \cdots $ then $\cup_{i=1}^{\infty} \langle a_i\rangle $ is an ideal and then using structure of $\Bbb Z[\sqrt{-5}]$ I can prove that it is a principal ideal.

Answer 1

Fact 1. If $\langle a\rangle\subseteq \langle b\rangle$, then $b|a$. This is clear, and holds in any ring.

Define $N\colon\mathbb{Z}[\sqrt{-5}]\to\mathbb{Z}$ by $N(a+b\sqrt{-5}) = a^2+5b^2$. Then $N$ is multiplicative, so if $a|b$, then $N(a)|N(b)$.

Fact 2. The only units of $\mathbb{Z}[\sqrt{-5}]$ are $1$ and $-1$. Indeed, if $a+b\sqrt{-5}$ is a unit, then $a^2+5b^2$ divides $N(1)=1$, hence $b=0$ and $a=\pm 1$.

Consequences. A nonzero element $a\in\mathbb{Z}[\sqrt{-5}]$ has only finitely many divisors.

Therefore, if $a\neq 0$, then there are only finitely many principal ideals that properly contain $a$, and so any ascending chain of ideals must stabilize.

(Fact 2 is technically unnecessary, but this shows that in fact there are only finitely many elements $b$ such that $(a)\subseteq (b)$, which simplifies the argument)